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Re: Order Isomorphic
Posted:
Jan 31, 2013 2:00 AM
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On Jan 30, 11:01 pm, William Elliot <ma...@panix.com> wrote:
> Is every infinite subset S of omega_0 with the inherited order,
> order isomorphic to omega_0?
Yes, of course. (And you don't need the subscript; omega without a
subscript means omega_0, the first infinite ordinal.)
> Yes. S is an ordinal, a denumerable ordinal.
> Let eta be the order type of S.
No, S is *not* an ordinal, unless S = omega. S is a subset of an
ordinal, so it's a well-ordered set, so it's isomorphic to an ordinal,
so it's order type is an ordinal.
> [. . .]
> Does the same reasoning hold to show that an uncountable subset
> of omega_1 with the inherited order is order isomorphic to omega_1.
Yes, of course. The ordinal omega_1 is the unique ordinal such that
(a) it';s uncountable, and (b) each of its proper initial segments is
countable. Any well-ordered set with those two properties is
isomorphic to omega_1. If S is an uncountable subset of omega_1, then
S (with the natural order) has those two properties.
More generally (and just as trivially), if kappa is an initial
ordinal, and if S is a subset of kappa which has the the same
cardinality as kappa, then S is order-isomorphic to kappa.
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